cos(7p/12) in surd form
By Anonymous on Monday, March 12, 2001 - 04:14 pm :
How do you evaluate something like this in surd form? cos(7p/12)=?
By Michael Doré (Md285) on Monday, March 12, 2001 - 04:29 pm :

You can use the double angle formula. We have:

cos(7p/6)=2cos2(7p/12)-1

(using cos2A=2cos2 A-1)

The left hand side is -cos(p/6)=-Ö3/2.

So

cos2(7p/12)=1/2-Ö3/4

Therefore (as cos(7p/12 < 0) we see the answer is:


-
Ö
 
1/2-Ö3/4
 

By Kerwin Hui (Kwkh2) on Monday, March 12, 2001 - 04:33 pm :
Another way is to note that 7p/12=p/3+p/4. So

cos(7p/12)=cos(p/4)cos(p/3)-sin(p/4)sin(p/3)=1/Ö2× 1/2-1/Ö2×Ö3/2=2-1.5(1-30.5)

Kerwin

By Anonymous on Monday, March 12, 2001 - 04:55 pm :

Thank you Michael, Kerwin.


Michael, how did you figure out that cos(7p/12) < 0?
By Barkley Bellinger (Bb246) on Monday, March 19, 2001 - 04:45 pm :
A previous anonymous questioner asked how Michael knew that cos(7p/12) < 0. If you know your cosine graph then you will know that cos(x) < 0 for p/2 < x < p, which 7p/12 satisfies.